Mathematics · Limits, Continuity and Differentiability

JEE Advanced 2019 — Paper 2 — Question 24

For $a \in \mathbb{R}|a|>1$ , let

\lim _{n \rightarrow \infty} \frac{1+\sqrt[3]{2}+\ldots+\sqrt[3]{n}}{\left.n^{7 / 3}-\frac{1}{(\text { an }+1)^{2}}+\frac{1}{(\text { an }+2)^{2}}+\ldots+\frac{1}{(\text { an }+n)^{2}}\right)}=54

Then the possible value(s) of a is/are

Option A:
8
Correct
Option B:
$-6$
Option C:
7
Option D:
$-9$
Correct

Answer: A, D

Step-by-step solution

\begin{aligned} & \lim _{n \rightarrow \infty} \frac{\sqrt[3]{1}+\sqrt[3]{2}+\ldots+\sqrt[3]{n}}{\left.n^{7 / 3} \frac{1}{-(x a+1)^{2}}+\frac{1}{(x a+2)^{2}}+\ldots+\frac{1}{(x a+n)^{2}}\right)}=54 \end{aligned}

\begin{aligned} & =\frac{\int_{0}^{1} x^{1 / 3} d x}{\int_{0}^{1} \frac{d x}{(a+x)^{2}}}=\frac{\left.\frac{3}{4} x^{4 / 3}\right|_{0} ^{1}}{-\frac{1}{-a+x} \|_{0}^{1}}=\frac{\frac{3}{4}}{-\frac{1}{a+1}-\frac{1}{a}}=54 \end{aligned}

\Rightarrow \quad \frac{3}{4 \frac{1}{a(a+1)}}=54 \Rightarrow a=8 \text { or } a=-9

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 2
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Evaluation of Limit of Functions